Homework Answers
Total number of pages = 4K = 4000
bits required to store 4000 pages = log2(4000) = 12 bits
bits required for 128 segments = log2(128) = 7 bits
number of pages in each segment = 4000/128 = 32 approx.
number of bits required to address 32 pages ie one segment = log2(32) = 5
Maximum size of each segment is when all 32 pages are present = 32 * 5 = 160 bits = 20 bytes
Total size of all segments = 128 * 20 = 2560 bytes = 2.5 KB
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